Answer:
11 hours
Step-by-step explanation:
Find how many hours the trip will take by dividing the number of miles by the rate in miles per hour:
660/60
= 11
So, the trip will take 11 hours
Do the following lengths form a right triangle?
Answer:
Yes
Step-by-step explanation:
The lengths of this right angle triangle (6, 8, 10) proves that the polygon is indeed a right angle triangle. This is because there are certain ratios to prove that a right angle triangle is indeed a right angle triangle. These are called the Pythagorean Triples . Some examples include; (3, 4, 5), (7, 24, 25) and (28, 45, 53). The Pythagorean Triple 3, 4, 5 can be scaled up to provide the triple 6, 8, 10, where the scale factor is 2.
reflectiion across y=x
9514 1404 393
Answer:
see attached
Step-by-step explanation:
The reflection across y=-x swaps the coordinates and negates both of them. The first-quadrant figure becomes a third-quadrant figure.
(x, y) ⇒ (-y, -x)
FLIGHT TO TOKYO TAKE 2 HOURS 20 MINUTES U ARRIVE AT 4:15PM WHICH TIME DID HE SET OFF
Answer: 1:55 PM
Step-by-step explanation:
Turn 4:15 to 24-hr clock system which is 1615hrs
16:15 - 02:20 = 1355hrs
Which term describes the ability to repay debt?
A
capacity
B.
creditworthiness
c. capability
Answer:
B. CreditworthinessStep-by-step explanation:
❣️(◍•Jess bregoli•◍)❣️#keep learning!!Answer:
B
Step-by-step explanation:
What is the solution to the HELP HELP equation m+4
m2
-?
m2-16
4-m
Om =-4
O m = -2
Om = 2.
Om = 4
Answer:
m = -2
Step-by-step explanation:
[tex]\frac{m}{m+4} + \frac{4}{4-m} = \frac{m^2}{m^2-16}\\\\\frac{m}{m+4} + \frac{4}{-(m-4)} = \frac{m^2}{m^2-16}\\\\[/tex] [tex][ \ ( 4 - m) = - (m -4) \ ][/tex]
[tex]\frac{m}{m+4} + \frac{-4}{(m-4)} = \frac{m^2}{m^2-16}\\\\[/tex] [tex][ \ \frac{a}{-b} = \frac{-a}{b}\ ][/tex]
[tex]\frac{m}{m+4} - \frac{4}{m-4} = \frac{m^2}{m^2-16}\\\\[/tex] [tex][ \ \frac{x}{y} + \frac{-a}{b} = \frac{x}{y} - \frac{a}{b} \ ][/tex]
[tex]\frac{m(m-4)-4(m+4)}{(m-4)(m+4)} = \frac{m^2}{m^2-16}\\\\\frac{m^2 -4m -4m-16}{(m^2-16)} = \frac{m^2}{m^2-16}\\\\[/tex] [tex][ \ T aking \ LCM \ and \ simplifying\ ][/tex]
[tex]m^2 - 8m -16 = m^2[/tex] [tex][ \ \frac{a}{b} = \frac{c}{b} => a = c \ ][/tex]
[tex]m^2 - m^2 - 8m - 16 = 0\\\\0 -8m = 16\\\\-8m = 16 \\\\m = \frac{16}{-8} = -2[/tex]
someone plz help me porfavor!!!!!
Answer:
c. y = ¼x - 2
Step-by-step explanation:
Find the slope (m) and y-intercept (b) then substitute the values into y = mx + b (slope-intercept form)
Slope = change in y/change in x
Using two points on the graph, (0, -2) and (4, -1):
Slope (m) = (-1 - (-2))/(4 - 0) = 1/4
m = ¼
y-intercept = the point where the line intercepts the y-axis = -2
b = -2
✔️To write the equation, substitute m = ¼ and b = -2 into y = mx + b:
y = ¼x - 2
what is the sum factor of 3600
Answer:
Step-by-step explanation:
Answer:
24
Step-by-step explanation:
Find the prime factorization of the number 3,600. Factor Tree.
2|3,600.
2|1,800.
2|900.
2|450
5|225
5|45
3|9
3|3
|1
Setup the equation for determining the number of factors or divisors.
3600=2x2x2x2x3x3x5
Sum factors=2+2+2+2+2+3+3+5=24
I need to find the equal expression to -m(2m+2n)+3mn+2m². Help please?
[tex]m(2m+2n)+3mn+2m^2\implies \stackrel{\textit{distributing}}{2m^2+2mn}+3mn+2m^2 \\\\\\ 2m^2+2m^2+2mn+3mn\implies \stackrel{\textit{adding like-terms}}{4m^2+5mn}[/tex]
What is the surface area of the regular pyramid below?
Compute the product AB by the definition of the product ofmatrices, where Ab1 and Ab2 are computed separately, and by therow-column rule for computing AB.
Matrix A= [2 -2]
[3 4]
[4 -3]
Matrix B =
[4 -1]
[-1 2]
Answer:
[tex]A * B = \left[\begin{array}{ccc}10&-6\\8&5\\19&-10\end{array}\right][/tex]
Step-by-step explanation:
Given
[tex]A =\left[\begin{array}{cc}2&-2\\3&4\\4&-3\end{array}\right][/tex]
[tex]B = \left[\begin{array}{cc}4&-1\\-1&2\end{array}\right][/tex]
Required
[tex]AB[/tex]
To do this, we simply multiply the rows of A by the column of B;
So, we have:
[tex]A * B = \left[\begin{array}{ccc}2*4 + -2*-1&2*-1+-2*2\\3*4+4*-1&3*-1+4*2\\4*4-3*-1&4*-1-3*2\end{array}\right][/tex]
[tex]A * B = \left[\begin{array}{ccc}10&-6\\8&5\\19&-10\end{array}\right][/tex]
Find the product with the exponent in simplest
form. Then, identify the values of x and y.
6
X
- 64
.
6
X
y =
DONE
Answer:
[tex]\displaystyle 8^\bigg{\frac{8}{3}}[/tex]
General Formulas and Concepts:
Pre-Algebra
Order of Operations: BPEMDAS
Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to RightAlgebra I
Exponential Rule [Powering]: [tex]\displaystyle (b^m)^n = b^{m \cdot n}[/tex]Step-by-step explanation:
Step 1: Define
Identify
[tex]\displaystyle \bigg(8^\bigg{\frac{2}{3}} \bigg)^4[/tex]
Step 2: Simplify
Exponential Rule [Powering]: [tex]\displaystyle 8^\bigg{\frac{2}{3} \cdot 4}[/tex]Multiply: [tex]\displaystyle 8^\bigg{\frac{8}{3}}[/tex]Find value of x of angles
Answer:
32
Step-by-step explanation:
52=x+20
52-20=x+20-20
32=x
Somebody please help asap
Answer:
B. [tex] 4x^2 + \frac{3}{2}x - 7 [/tex]
Step-by-step explanation:
[tex] f(x) = \frac{x}{2} - 3 [/tex]
[tex] g(x) = 4x^2 + x - 4 [/tex]
(f + g)(x) = f(x) - g(x)
= [tex] \frac{x}{2} - 3 + 4x^2 + x - 4 [/tex]
Add like terms
[tex] = 4x^2 + \frac{x}{2} + x - 3 - 4 [/tex]
[tex] = 4x^2 + \frac{3x}{2} - 7 [/tex]
[tex] = 4x^2 + \frac{3}{2}x - 7 [/tex]
Annual windstorm losses, X and Y, in two different regions are independent, and each is uniformly distributed on the interval [0, 10]. Calculate the covariance of X and Y, given that X+ Y < 10.
Answer:
[tex]Cov(X,Y) = -\frac{ 25}{9}[/tex]
Step-by-step explanation:
Given
[tex]Interval =[0,10][/tex]
[tex]X + Y < 10[/tex]
Required
[tex]Cov(X,Y)[/tex]
First, we calculate the joint distribution of X and Y
Plot [tex]X + Y < 10[/tex]
So, the joint pdf is:
[tex]f(X,Y) = \frac{1}{Area}[/tex] --- i.e. the area of the shaded region
The shaded area is a triangle that has: height = 10; width = 10
So, we have:
[tex]f(X,Y) = \frac{1}{0.5 * 10 * 10}[/tex]
[tex]f(X,Y) = \frac{1}{50}[/tex]
[tex]Cov(X,Y)[/tex] is calculated as:
[tex]Cov(X,Y) = E(XY) - E(X) \cdot E(Y)[/tex]
Calculate E(XY)
[tex]E(XY) =\int\limits^X_0 {\int\limits^Y_0 {\frac{XY}{50}} \, dY} \, dX[/tex]
[tex]X + Y < 10[/tex]
Make Y the subject
[tex]Y < 10 - X[/tex]
So, we have:
[tex]E(XY) =\int\limits^{10}_0 {\int\limits^{10 - X}_0 {\frac{XY}{50}} \, dY} \, dX[/tex]
Rewrite as:
[tex]E(XY) =\frac{1}{50}\int\limits^{10}_0 {\int\limits^{10 - X}_0 {XY}} \, dY} \, dX[/tex]
Integrate Y
[tex]E(XY) =\frac{1}{50}\int\limits^{10}_0 {\frac{XY^2}{2}}} }|\limits^{10 - X}_0 \, dX[/tex]
Expand
[tex]E(XY) =\frac{1}{50}\int\limits^{10}_0 {\frac{X(10 - X)^2}{2} - \frac{X(0)^2}{2}}} }\ dX[/tex]
[tex]E(XY) =\frac{1}{50}\int\limits^{10}_0 {\frac{X(10 - X)^2}{2}}} }\ dX[/tex]
Rewrite as:
[tex]E(XY) =\frac{1}{100}\int\limits^{10}_0 X(10 - X)^2\ dX[/tex]
Expand
[tex]E(XY) =\frac{1}{100}\int\limits^{10}_0 X*(100 - 20X + X^2)\ dX[/tex]
[tex]E(XY) =\frac{1}{100}\int\limits^{10}_0 100X - 20X^2 + X^3\ dX[/tex]
Integrate
[tex]E(XY) =\frac{1}{100} [\frac{100X^2}{2} - \frac{20X^3}{3} + \frac{X^4}{4}]|\limits^{10}_0[/tex]
Expand
[tex]E(XY) =\frac{1}{100} ([\frac{100*10^2}{2} - \frac{20*10^3}{3} + \frac{10^4}{4}] - [\frac{100*0^2}{2} - \frac{20*0^3}{3} + \frac{0^4}{4}])[/tex]
[tex]E(XY) =\frac{1}{100} ([\frac{10000}{2} - \frac{20000}{3} + \frac{10000}{4}] - 0)[/tex]
[tex]E(XY) =\frac{1}{100} ([5000 - \frac{20000}{3} + 2500])[/tex]
[tex]E(XY) =50 - \frac{200}{3} + 25[/tex]
Take LCM
[tex]E(XY) = \frac{150-200+75}{3}[/tex]
[tex]E(XY) = \frac{25}{3}[/tex]
Calculate E(X)
[tex]E(X) =\int\limits^{10}_0 {\int\limits^{10 - X}_0 {\frac{X}{50}}} \, dY} \, dX[/tex]
Rewrite as:
[tex]E(X) =\frac{1}{50}\int\limits^{10}_0 {\int\limits^{10 - X}_0 {X}} \, dY} \, dX[/tex]
Integrate Y
[tex]E(X) =\frac{1}{50}\int\limits^{10}_0 { (X*Y)|\limits^{10 - X}_0 \, dX[/tex]
Expand
[tex]E(X) =\frac{1}{50}\int\limits^{10}_0 ( [X*(10 - X)] - [X * 0])\ dX[/tex]
[tex]E(X) =\frac{1}{50}\int\limits^{10}_0 ( [X*(10 - X)]\ dX[/tex]
[tex]E(X) =\frac{1}{50}\int\limits^{10}_0 10X - X^2\ dX[/tex]
Integrate
[tex]E(X) =\frac{1}{50}(5X^2 - \frac{1}{3}X^3)|\limits^{10}_0[/tex]
Expand
[tex]E(X) =\frac{1}{50}[(5*10^2 - \frac{1}{3}*10^3)-(5*0^2 - \frac{1}{3}*0^3)][/tex]
[tex]E(X) =\frac{1}{50}[5*100 - \frac{1}{3}*10^3][/tex]
[tex]E(X) =\frac{1}{50}[500 - \frac{1000}{3}][/tex]
[tex]E(X) = 10- \frac{20}{3}[/tex]
Take LCM
[tex]E(X) = \frac{30-20}{3}[/tex]
[tex]E(X) = \frac{10}{3}[/tex]
Calculate E(Y)
[tex]E(Y) =\int\limits^{10}_0 {\int\limits^{10 - X}_0 {\frac{Y}{50}}} \, dY} \, dX[/tex]
Rewrite as:
[tex]E(Y) =\frac{1}{50}\int\limits^{10}_0 {\int\limits^{10 - X}_0 {Y}} \, dY} \, dX[/tex]
Integrate Y
[tex]E(Y) =\frac{1}{50}\int\limits^{10}_0 { (\frac{Y^2}{2})|\limits^{10 - X}_0 \, dX[/tex]
Expand
[tex]E(Y) =\frac{1}{50}\int\limits^{10}_0 ( [\frac{(10 - X)^2}{2}] - [\frac{(0)^2}{2}])\ dX[/tex]
[tex]E(Y) =\frac{1}{50}\int\limits^{10}_0 ( [\frac{(10 - X)^2}{2}] )\ dX[/tex]
[tex]E(Y) =\frac{1}{50}\int\limits^{10}_0 [\frac{100 - 20X + X^2}{2}] \ dX[/tex]
Rewrite as:
[tex]E(Y) =\frac{1}{100}\int\limits^{10}_0 [100 - 20X + X^2] \ dX[/tex]
Integrate
[tex]E(Y) =\frac{1}{100}( [100X - 10X^2 + \frac{1}{3}X^3]|\limits^{10}_0)[/tex]
Expand
[tex]E(Y) =\frac{1}{100}( [100*10 - 10*10^2 + \frac{1}{3}*10^3] -[100*0 - 10*0^2 + \frac{1}{3}*0^3] )[/tex]
[tex]E(Y) =\frac{1}{100}[100*10 - 10*10^2 + \frac{1}{3}*10^3][/tex]
[tex]E(Y) =10 - 10 + \frac{1}{3}*10[/tex]
[tex]E(Y) =\frac{10}{3}[/tex]
Recall that:
[tex]Cov(X,Y) = E(XY) - E(X) \cdot E(Y)[/tex]
[tex]Cov(X,Y) = \frac{25}{3} - \frac{10}{3}*\frac{10}{3}[/tex]
[tex]Cov(X,Y) = \frac{25}{3} - \frac{100}{9}[/tex]
Take LCM
[tex]Cov(X,Y) = \frac{75- 100}{9}[/tex]
[tex]Cov(X,Y) = -\frac{ 25}{9}[/tex]
The profits in hundreds of dollars, P(c), that a company can make from a product is modeled by a function of the price, c, they charge for the product: P(c) = –20c2 + 320c + 5,120. What is the maximum profit the company can make from the product?
Answer:
6400
Step-by-step explanation:
Given the profit function ;
P(c) = –20c2 + 320c + 5,120
The maximum value is given by :
f(h) ; where, h = - b /2a
From P(C) ; a = - 20 ; b = 320
h = - b / 2a = - 320 / 2(-20) = - 320 / 40 = 8
c = h
P(8) = –20(8)² + 320(8) + 5,120
P(8) = - 1280 + 2560 + 5120
= 6400
Answer:
B.) $640,000
Step-by-step explanation:
Given the functions:
g(n) = 3n - 5
f(n) = n2 + 50
Find:
(g+f)(8)
Answer:
[tex](g + f)(8) =133[/tex]
Step-by-step explanation:
Given
[tex]g(n) = 3n - 5[/tex]
[tex]f(n) = n^2 + 50[/tex]
Required
[tex](g + f)(8)[/tex]
This is calculated as:
[tex](g + f)(n) =g(n) + f(n)[/tex]
So, we have:
[tex](g + f)(n) =3n - 5 + n^2 +50[/tex]
[tex]Substitute[/tex] 8 for n
[tex](g + f)(8) =3*8 - 5 + 8^2 +50[/tex]
[tex](g + f)(8) =24 - 5 + 64 +50[/tex]
[tex](g + f)(8) =133[/tex]
The length of a rectangle should be 9 meters longer than 7 times the width. If the length must be
between 93 and 163 meters long, what are the restrictions for the width, p?
Write the solution set as an algebraic inequality solved for the variable.
Answer:
If we define W as the width:
12m ≤ W ≤ 22m
Step-by-step explanation:
We have a rectangle with length L and width W.
We know that:
"The length of a rectangle should be 9 meters longer than 7 times the width"
Then:
L = 9m + 7*W
We also know that the length must be between 93 and 163 meters long, so:
93m ≤ L ≤ 163m
Now we want to find the restrictions for the width W.
We start with:
93m ≤ L ≤ 163m
Now we know that L = 9m + 7*W, then we can replace that in the above inequality:
93m ≤ 9m + 7*W ≤ 163m
Now we need to isolate W.
First, we can subtract 9m in the 3 sides of the inequality
93m - 9m ≤ 9m + 7*W -9m ≤ 163m -9m
84m ≤ 7*W ≤ 154m
Now we can divide by 7 in the 3 sides, so we get:
84m/7 ≤ 7*W/7 ≤ 154m/7
12m ≤ W ≤ 22m
Then we can conclude that the width is between 12 and 22 meters long.
the second difference 4;x;8;y;20;.... is 2
9514 1404 393
Answer:
x = 5y = 13Step-by-step explanation:
First differences are ...
x -4, 8 -x, y -8, 20 -y
Then second differences are ...
(8 -x) -(x -4)
(y -8) -(8 -x)
(20 -y) -(y -8)
Each of these is said to be 2, so we have ...
12 -2x = 2 ⇒ x = 5
28 -2y = 2 ⇒ y = 13
And an equation we can use to check:
x +y -16 = 2 ⇒ 5 +13 -16 = 2 . . . . true
_____
The explicit formula for the sequence is ...
an = n^2 -2n +5
The box plots show the weights, in pounds, of the dogs in two different animal shelters.
Weights of Dogs in Shelter A
2 box plots. The number line goes from 6 to 30. For the weights of dogs in shelter A, the whiskers range from 8 to 30, and the box ranges from 17 to 28. A line divides the box at 21. For shelter B, the whiskers range from 10 to 28, and the box ranges from 16 to 20. A line divides the box at 18.
Weights of Dogs in Shelter B
Which is true of the data in the box plots? Select three choices.
The median weight for shelter A is greater than that for shelter B.
The median weight for shelter B is greater than that for shelter A.
The data for shelter A are a symmetric data set.
The data for shelter B are a symmetric data set.
The interquartile range of shelter A is greater than the interquartile range of shelter B.
Answer:
The median weight for shelter A is greater than that for shelter B.
The data for shelter B are a symmetric data set.
The interquartile range of shelter A is greater than the interquartile range of shelter B.
Step-by-step explanation:
The median weight for shelter A is greater than that for shelter B.
The median of A = 21 and the median of B = 18 true
The median weight for shelter B is greater than that for shelter A.
The median of A = 21 and the median of B = 18 false
The data for shelter A are a symmetric data set.
False, looking at the box it is not symmetric
The data for shelter B are a symmetric data set.
true, looking at the box it is symmetric
The interquartile range of shelter A is greater than the interquartile range of shelter B.
IQR = 28 - 17 = 11 for A
IQR for B = 20 -16 = 4 True
A trailer is 22 feet long. 9 feet wide,
and 7 feet high. What is the volume of
the trailer?
Answer:
1386
Step-by-step explanation:
22 × 9 × 7 = 1386 cubic feet
In The Diagram, P, Q, R, Are Points On The Circle With Centre 0 And Diameter 14cm. Angle PQR=35 Degree. Find, Correct To One Decimal Place A. The Length Of The Minor Are PQ; B. The Chord PQ.
This question hasn't been solved yet
Ask an expert
in the diagram, P, Q, R, are points on the circle with centre 0 and diameter 14cm. angle PQR=35 degree. find, correct to one decimal place
a. The length of the minor are PQ;
B. The chord PQ.
Answer:
a
Step-by-step explanation:
because the diagram the length decimal place degree
Which of the following is NOT a solution to the linear equation y=3x+2?
Select the correct answer below:
(1,5)
(2,8)
(3,10)
(4,14)
Answer:
(3,10)
Step-by-step explanation:
When x is 3
[tex]{ \bf{y = 3x + 2}} \\ { \bf{y = (3 \times 3) + 2}} \\ { \bf{y = 11}}[/tex]
y is 11, and this is invalid because it is not at accord.
The correct coordinates which is NOT a solution to the linear equation
y = 3x+2 is,
⇒ (3, 10)
What is Equation of line?The equation of line in point-slope form passing through the points
(x₁ , y₁) and (x₂, y₂) with slope m is defined as;
⇒ y - y₁ = m (x - x₁)
Where, m = (y₂ - y₁) / (x₂ - x₁)
Given that;
Linear equation is,
⇒ y = 3x + 2
We know that;
The solution of linear equation is satisfy the eqaution.
Hence, We can check as;
⇒ y = 3x + 2
Put x = 1. y = 5
⇒ 5 = 3 × 1 + 2
⇒ 5 = 5
Thus, It is solution of linear equation.
⇒ y = 3x + 2
Put x = 2. y = 8
⇒ 8 = 3 × 2 + 2
⇒ 8 = 8
Thus, It is solution of linear equation.
⇒ y = 3x + 2
Put x = 3, y = 10
⇒ 10 = 3 × 3 + 2
⇒ 10 ≠ 11
Thus, It is not solution of linear equation.
Thus, The correct coordinates which is NOT a solution to the linear equation y = 3x+2 is,
⇒ (3, 10)
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You measure 37 dogs' weights, and find they have a mean weight of 69 ounces. Assume the population standard deviation is 9.2 ounces. Based on this, construct a 90% confidence interval for the true population mean dog weight.Give your answers as decimals, to two places_______ ± ________ ounces (Enter your answer as an integer or decimal number. Examples: 3, -4, 5.5172 ; Enter DNE for Does Not Exist, oo for Infinity)
Solution :
Given :
mean weight of the dogs, [tex]$\overline x$[/tex] = 69
Number of dogs, n = 37
Standard deviation, σ = 9.2
Confidence interval = 90%
At 90% confidence interval for the true population mean dog weight is given by :
[tex]$= \overline x \pm \frac{\sigma}{\sqrt n} \times z_{0.05}$[/tex]
[tex]$= 69\pm \frac{9.2}{\sqrt_{37}}} \times 1.64485$[/tex]
[tex]$=69 \pm 2.487799$[/tex]
= (66.5122, 71.4878)
If 21% of kindergarten children are afraid of monsters, how many out of
each 100 are afraid?
Answer:
The appropriate answer is "21".
Step-by-step explanation:
Given:
Afraid percentage,
p = 21%
or,
= 0.21
Sample size,
n = 100
As we know,
⇒ [tex]X=np[/tex]
By putting the values, we get
[tex]=0.21\times 100[/tex]
[tex]=21[/tex]
2x-5y=22n y=3x-7 Use substitution to solve the system.
Answer:
x = 1 , y = -4
Step-by-step explanation:
2x - 5y = 22 ------- ( 1 )
y = 3x - 7 ------- ( 2 )
Substitute ( 2 ) in ( 1 ) :
2x - 5 (3x - 7) = 22
2x - 15x + 35 = 22
- 13x = 22 - 35
- 13x = - 13
x = 1
Substitute x in ( 1 ) :
2x - 5y = 22
2 ( 1 ) - 5y = 22
- 5y = 22 - 2
-5y = 20
y = - 4
The human resource department at a certain company wants to conduct a survey regarding worker benefits. The department has an alphabetical list of all 4247 employees at the company and wants to conduct a systematic sample of size 40.
Required:
a. What is k?
b. Determine the individuals who will be administered the survey. Randomly select a number from 1 to k. Suppose that we randomly select 19. Starting with the first individual selected, the individuals in the survey will be: _________
Answer:
a). 106
b). See Explanation
Step-by-step explanation:
According to the Question,
Given, The human resource department at a certain company wants to conduct a survey regarding worker benefits. The department has an alphabetical list of all 4247 employees at the company and wants to conduct a systematic sample of size 40.
a). Thus, the Required Value of K is 4247/40 = 106.175 ≈ 106
b). The individuals who will be administered the survey. Randomly select a number from 1 to k. Suppose that we randomly select 19. Starting with the first individual selected, the individuals in the survey will be
19(19+106) = 125(125+106) = 231337443549655761867973107911851291139715031609171518211927203321392245235124572563266927752881298730933199330534113517362337293835394140474153Wayne is picking out some movies to rent, and he is primarily interested in dramas and horror films. He has narrowed down his selections to 7 dramas and 16 horror films. How many different combinations of 3 movies can he rent if he wants at least two dramas
Answer:
The number of selections is 49.
Step-by-step explanation:
drama = 7
horror films = 16
Select 3 movies at least two dramas
For 2 drama and 1 horror film
(3 C 2) x (16 C 1) = 48
For 3 drama
(3 C 3) = 1
So, total number of selections is 48 + 1 = 49.
José tiene 30 años menos que su padre y 27 más que su hijo. entre los 3 suman 135 años ¿ cuántos años tiene cada uno?
Answer:
44 años tiene jose . el padre 74 y el hijo 17 años.
Step-by-step explanation:
A candy distributor needs to mix a 40% fat-content chocolate with a 60% fat-content chocolate to create 150 kilograms of a 52% fat-content chocolate. How many kilograms of each kind of chocolate must they use?
Answer:
60 kg of 40% and 90 kg of 60%
Step-by-step explanation:
Let the amount of 40% chocolate be x.
Let the amount of 60% chocolate be y.
x + y = 150
0.4x + 0.6y = 0.52 * 150
x + y = 150
4x + 6y = 780
-4x - 4y = -600
(+) 4x + 6y = 780
--------------------------
2y = 180
y = 90
x + y = 150
x + 90 = 150
x = 60
Answer: 60 kg of 40% and 90 kg of 60%
What is a Parrel line?
Answer:
parrel line never meet
